Given a string s and a non-empty string p, find all the start indices of p's anagrams in s.
Strings consists of lowercase English letters only and the length of both strings s and p will not be larger than 20,100.
The order of output does not matter.
Example 1:
Input:
s: "cbaebabacd" p: "abc"

Output:
[0, 6]

Explanation:
The substring with start index = 0 is "cba", which is an anagram of "abc".
The substring with start index = 6 is "bac", which is an anagram of "abc".
Example 2:

Input:
s: "abab" p: "ab"

Output:
[0, 1, 2]

Explanation:
The substring with start index = 0 is "ab", which is an anagram of "ab".
The substring with start index = 1 is "ba", which is an anagram of "ab".
The substring with start index = 2 is "ab", which is an anagram of "ab".

思路:

用一个滑动窗口,随时判断窗口里面的字符是否与p字符串相符合anagram。

vector<int> findAnagrams(string s, string p) {
vector<int> pv(,), sv(,), res;
if(s.size() < p.size())
return res;
for(int i = ; i < p.size(); ++i)
{
++pv[p[i]];
++sv[s[i]];
}
if(pv == sv)
res.push_back();
for(int i = p.size(); i < s.size(); ++i)
{
++sv[s[i]];
--sv[s[i-p.size()]];
if(pv == sv)
res.push_back(i-p.size()+);
}
return res;
}

参考:

https://discuss.leetcode.com/topic/64390/c-o-n-sliding-window-concise-solution-with-explanation

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